Question

THIS IS THE SECOND TIME IM PUTTING THIS HERE PLS ITS URGENT I STRUGGLE WITH CALCULUS THANK YOU

Answer 1

$v(t)=t^2-9t+18$

1. The velocity is the rate of change of the particle's position over time. The average velocity is then the average rate of change of the particle's position. If $s(t)$ denotes the particle's position at time $t$, then its velocity at time $t$ satisfies $s'(t)=v(t)$. So to get the average velocity, we need to integrate:

$\text{average vel.}=\dfrac{s(8)-s(0)}{8-0}=\displaystyle\frac18\int_0^8v(t)\,\mathrm dt$

which follows from the fundamental theorem of calculus. If $s'(t)=v(t)$, then

$\displaystyle\int_a^bv(t)\,\mathrm dt=s(b)-s(a)$

We get an average velocity of

$\displaystyle\frac18\left(\frac{t^3}3-\frac{9t^2}2+18t\right)\bigg|_{t=0}^{t=8}=\frac{10}3$

(in meters/second).

2. The instantaneous velocity can be found by simply evaluating the velocity function at the given time. We get $v(5)=-2$ m/s. Speed is the same as velocity, but we don't account for direction, meaning we just take the magnitude, given by the absolute value of the velocity. So the particle's speed is $|v(5)|=2[tex] m/s. 3. We're using the convention that moving "to the right of zero" corresponds to having a positive velocity. This happens with [tex]v(t)=t^2-9t+18=(t-3)(t-6)>0$

which is satisfied when both $t-3$ and $t-6$ are positive, or both are negative. This occurs when $t<3$ or when $t>6$.

4. The particle is moving faster or slower when the rate of change of its velocity is positive or negative, respectively. So we have to check the derivative of $v(t)$ (the acceleration), find its critical points, then examine where the derivative changes sign.

$v'(t)=2t-9=0\implies t=\dfrac92$

a. When $t>\dfrac92$, we have $v'(t)>0$, so the particle is speeding up.

b. When $t<\dfrac92$, we have $v'(t)<0$, so the particle is slowing down.

5. The total distance traveled can be determined integrating the particle's speed function. When we do so, we're essentially chunking the total time interval into smaller subintervals, checking how much distance is traveled over each subinterval by multiplying the average speed over each subinterval by the length (duration) of each subinterval, and adding all these subdistances together.

So we compute:

$\displaystyle\int_0^8|v(t)|\,\mathrm dt=\int_0^8|t^2-9t+18|\,\mathrm dt$

In the analysis from part (3), we know $v(t)<0$ when $3<t<6$, so it must be non-negative elsewhere. What this means is we have

$\displaystyle\int_0^8|v(t)|\,\mathrm dt=\int_0^3v(t)\,\mathrm dt+\int_3^6(-v(t))\,\mathrm dt+\int_6^8v(t)\,\mathrm dt$

because $|v(t)|=-v(t)$ whenever $v(t)<0$, and $|v(t)|=v(t)$ whenever $v(t)>0$. (This is just the definition of the absolute value function.)

Computing each integral, we get a total distance traveled of $\dfrac{107}3$ meters.